当我使用第一个查询计算第二个查询的值大于或等于 2 时，我正在显示输出，其中第二个查询计划编号等于第一个查询计划编号。
我收到错误的行:
if ($x>=2) {.'<button onclick="CreateRevision('.$row['id'].')" type="button" class="btn btn-primary"><span class="fa fa-copy"></span></button>'.} else{}.' .
我刚刚收到这条消息syntax error, unexpected'.'
我的整个代码是这样的:
<?php
//open connection to mysql db
include("db_connection.php");
$data = '
<table id="tb_jobsched" class="display nowrap table table-bordered" width="100%" cellspacing="0">
<thead>
<tr>
<th width="3%">#</th>
<th width="10%">Schedule Number</th>
<th width="20%">Item Name</th>
<th width="10%">Output</th>
<th width="10%">Date Created</th>
<th width="10%">Expected Date</th>
<th width="10%">Prepared By</th>
<th width="7%">Status</th>
<th width="7%"></th>
</tr>
</thead>
<tfoot>
<tr>
<th width="3%">#</th>
<th width="10%">Schedule Number</th>
<th width="20%">Item Name</th>
<th width="10%">Output</th>
<th width="10%">Date Created</th>
<th width="10%">Expected Date</th>
<th width="10%">Prepared By</th>
<th width="7%">Status</th>
<th width="7%"></th>
</tr>
</tfoot>';
$query = "SELECT * FROM prepressjobs WHERE pj_deleted = 0 ORDER BY pj_datecreated DESC";
if (!$result = mysqli_query($db,$query)) {
exit(mysqli_error());
}
// if query results contains rows then featch those rows
if(mysqli_num_rows($result) > 0)
{
$number = 1;
while($row = mysqli_fetch_assoc($result))
{
$data .= '<tr>
<td>'.$number.'</td>
<td>'.$row['pj_schednum'].'</td>
<td>'.$row['pj_itemname'].'</td>
<td>'.$row['pj_output'].'</td>
<td>'.$row['pj_datecreated'].'</td>
<td>'.$row['pj_expectedfinished'].'</td>
<td>'.$row['pj_preparedby'].'</td>
<td>'.$row['pj_status'].'</td>
<td>
<div class="btn-group" role="group" aria-label="Button group with nested dropdown">
<button onclick="GetJobDetails('.$row['id'].')" type="button" class="btn btn-primary"><span class="fa fa-gear"></span></button>
'.
$querya = "SELECT COUNT(rm_schednum) AS jobrevision FROM jobrevisions WHERE rm_schednum ='{$row['pj_schednum']}'";
$resulta= mysqli_query($db, $querya);
$rowa = mysqli_fetch_assoc($resulta);
$x=$rowa['jobrevision'];
if ($x>=2) {.'<button onclick="CreateRevision('.$row['id'].')" type="button" class="btn btn-primary"><span class="fa fa-copy"></span></button>'.} else{}.'
</div>
</td>
</tr>';
$number++;
}
}
else
{
// records now found
$data .= '<tr><td colspan="8">Records not found!</td></tr>';
}
$data .= '</table>';
echo $data;
?>
以下解决方案是什么？
最佳答案
<?php
<?php
//open connection to mysql db
include("db_connection.php");
$data = '
<table id="tb_jobsched" class="display nowrap table table-bordered" width="100%" cellspacing="0">
<thead>
<tr>
<th width="3%">#</th>
<th width="10%">Schedule Number</th>
<th width="20%">Item Name</th>
<th width="10%">Output</th>
<th width="10%">Date Created</th>
<th width="10%">Expected Date</th>
<th width="10%">Prepared By</th>
<th width="7%">Status</th>
<th width="7%"></th>
</tr>
</thead>
<tfoot>
<tr>
<th width="3%">#</th>
<th width="10%">Schedule Number</th>
<th width="20%">Item Name</th>
<th width="10%">Output</th>
<th width="10%">Date Created</th>
<th width="10%">Expected Date</th>
<th width="10%">Prepared By</th>
<th width="7%">Status</th>
<th width="7%"></th>
</tr>
</tfoot>';
$query = "SELECT * FROM prepressjobs WHERE pj_deleted = 0 ORDER BY pj_datecreated DESC";
if (!$result = mysqli_query($db,$query)) {
exit(mysqli_error());
}
// if query results contains rows then featch those rows
if(mysqli_num_rows($result) > 0)
{
$number = 1;
while($row = mysqli_fetch_assoc($result))
{
$data .= "<tr>
<td>{$number}</td>
<td>{$row['pj_schednum']}</td>
<td>{$row['pj_itemname']}</td>
<td>{$row['pj_output']}</td>
<td>{$row['pj_datecreated']}</td>
<td>{$row['pj_expectedfinished']}</td>
<td>{$row['pj_preparedby']}</td>
<td>{$row['pj_status']}</td>
<td>
<div class='btn-group' role='group' aria-label='Button group with nested dropdown'>
<button onclick='GetJobDetails({$row['id']})' type='button' class='btn btn-primary'><span class='fa fa-gear'></span></button>
";
$querya = "SELECT COUNT(rm_schednum) AS jobrevision FROM jobrevisions WHERE rm_schednum ='{$row['pj_schednum']}'";
$resulta= mysqli_query($db, $querya);
$rowa = mysqli_fetch_assoc($resulta);
$x = $rowa['jobrevision'];
if ($x>=2) {
$data .= "<button onclick='CreateRevision({$row['id']})' type='button' class='btn btn-primary'><span class='fa fa-copy'></span></button>";
} else{}
$data .= "</div>
</td>
</tr>";
$number++;
}
}
else
{
// records now found
$data .= "<tr><td colspan='8'>Records not found!</td></tr>";
}
$data .= "</table>";
echo $data;
?>
关于php - 在另一个查询中使用 if 和 using query 时出现语法错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/45503215/